# The general iterative methods for equilibrium problems and fixed point problems of a countable family of nonexpansive mappings in Hilbert spaces

## Abstract

In this paper, the researcher introduces the general iterative scheme for finding a common element of the set of equilibrium problems and fixed point problems of a countable family of nonexpansive mappings in Hilbert spaces. The results presented in this paper improve and extend the corresponding results announced by many others.

MSC:47H10, 47H09.

## 1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. Let C be a nonempty closed and convex subset of H, and let $T:C\to C$ be a nonlinear mapping. In this paper, we use $F\left(T\right)$ to denote the fixed point set of T.

Recall the following definitions.

1. (1)

The mapping T is said to be nonexpansive if

$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
(1.1)

Further, let F be a bifunction from $C×C$ into , where is the set of real numbers. The so-called equilibrium problem for $F:C×C\to \mathbb{R}$ is to find $y\in C$ such that

$F\left(y,u\right)\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C.$
(1.2)

The set of solutions of (1.2) is denoted by $EP\left(F\right)$. Given a mapping $A:C\to H$, let $F\left(y,u\right)=〈Ay,u-y〉$ for all $y,u\in C$. Then $z\in EP\left(F\right)$ if and only if $〈Az,u-z〉\ge 0$ for all $u\in C$. Numerous problems in physics, optimization and economics reduce to finding a solution of (1.2).

1. (2)

The mappings ${\left\{{T}_{n}\right\}}_{n\in \mathbb{N}}$ are said to be a family of nonexpansive mappings from H into itself if

$\parallel {T}_{n}x-{T}_{n}y\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in H,$
(1.3)

and denoted by $F\left({T}_{n}\right)=\left\{x\in H:{T}_{n}x=x\right\}$ is the fixed point set of ${T}_{n}$. Finding an optimal point in ${\bigcap }_{n\in \mathbb{N}}F\left({T}_{n}\right)$ of the fixed point sets of each mapping is a matter of interest in various branches of science.

Recently, many authors considered the iterative methods for finding a common element of the set of solutions to problem (1.2) and of the set of fixed points of nonexpansive mappings; see, for example, [1, 2] and the references therein.

Next, let $A:C\to H$ be a nonlinear mapping. We recall the following definitions.

1. (3)

A is said to be monotone if

$〈Ax-Ay,x-y〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
2. (4)

A is said to be strongly monotone if there exists a constant $\alpha >0$ such that

$〈Ax-Ay,x-y〉\ge \alpha {\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

In such a case, A is said to be α-strongly monotone.

1. (5)

A is said to be inverse-strongly monotone if there exists a constant $\alpha >0$ such that

$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

In such a case, A is said to be α-inverse-strongly monotone.

The classical variational inequality is to find $u\in C$ such that

$〈Au,v-u〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }v\in C.$
(1.4)

In this paper, we use $VI\left(C,A\right)$ to denote the set of solutions to problem (1.4). One can easily see that the variational inequality problem is equivalent to a fixed point problem. $u\in C$ is a solution to problem (1.4) if and only if u is a fixed point of the mapping ${P}_{C}\left(I-\lambda \right)T$, where $\lambda >0$ is a constant.

The variational inequality has been widely studied in the literature; see, for example, the work of Plubtieng and Punpaeng  and the references therein.

Recently, Ceng et al.  considered an iterative method for the system of variational inequalities (1.4). They got a strongly convergence theorem for problem (1.4) and a fixed point problem for a single nonexpansive mapping; see  for more details.

On the other hand, Moudafi  introduced the viscosity approximation method for nonexpansive mappings (see  for further developments in both Hilbert and Banach spaces).

A mapping $f:C\to C$ is called α-contractive if there exists a constant $\alpha \in \left(0,1\right)$ such that

$\parallel f\left(x\right)-f\left(y\right)\parallel \le \alpha \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
(1.5)

Let f be a contraction on C. Starting with an arbitrary initial ${x}_{1}\in C$, define a sequence $\left\{{x}_{n}\right\}$ recursively by

${x}_{n+1}=\left(1-{\sigma }_{n}\right)T{x}_{n}+{\sigma }_{n}f\left({x}_{n}\right),\phantom{\rule{1em}{0ex}}n\ge 0,$
(1.6)

where $\left\{{\sigma }_{n}\right\}$ is a sequence in $\left(0,1\right)$. It is proved [5, 6] that under certain appropriate conditions imposed on $\left\{{\sigma }_{n}\right\}$, the sequence $\left\{{x}_{n}\right\}$ generated by (1.6) strongly converges to the unique solution q in C of the variational inequality

$〈\left(I-f\right)q,p-q〉\ge 0,\phantom{\rule{1em}{0ex}}p\in C.$

Let A be a strongly positive linear bounded operator on a Hilbert space H with a constant $\overline{\gamma }$; that is, there exists $\overline{\gamma }>0$ such that

$〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H.$
(1.7)

Recently, Marino and Xu  introduced the following general iterative method:

${x}_{n+1}=\left(I-{\alpha }_{n}A\right)T{x}_{n}+{\alpha }_{n}\gamma f\left({x}_{n}\right),\phantom{\rule{1em}{0ex}}n\ge 0,$
(1.8)

where A is a strongly positive bounded linear operator on H. They proved that if the sequence $\left\{{\alpha }_{n}\right\}$ of parameters satisfies appropriate conditions, then the sequence $\left\{{x}_{n}\right\}$ generated by (1.8) converges strongly to the unique solution of the variational inequality

$〈\left(A-\gamma f\right){x}^{\ast },x-{x}^{\ast }〉\ge 0,\phantom{\rule{1em}{0ex}}x\in C,$
(1.9)

which is the optimality condition for the minimization problem

$\underset{x\in C}{min}\frac{1}{2}〈Ax,x〉-h\left(x\right),$

where h is a potential function for γf (i.e., ${h}^{\prime }\left(x\right)=\gamma f\left(x\right)$ for $x\in H$).

In 2007, Takahashi and Takahashi  introduced an iterative scheme by the viscosity approximation method for finding a common element of the set of solutions (1.2) and the set of fixed points of a nonexpansive mapping in Hilbert spaces. Let $S:C\to H$ be a nonexpansive mapping. Starting with arbitrary initial ${x}_{1}\in H$, define sequences $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ recursively by

$\left\{\begin{array}{c}F\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,\hfill \\ {x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{u}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.\hfill \end{array}$
(1.10)

They proved that under certain appropriate conditions imposed on $\left\{{\alpha }_{n}\right\}$ and $\left\{{r}_{n}\right\}$, the sequences $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ converge strongly to $z\in F\left(S\right)\cap EP\left(F\right)$, where $z={P}_{F\left(S\right)\cap EP\left(F\right)}f\left(z\right)$.

Next, Plubtieng and Punpaeng,  introduced an iterative scheme by the general iterative method for finding a common element of the set of solutions (1.2) and the set of fixed points of nonexpansive mappings in Hilbert spaces.

Let $S:H\to H$ be a nonexpansive mapping. Starting with an arbitrary ${x}_{1}\in H$, define sequences $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ by

$\left\{\begin{array}{c}F\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,\hfill \\ {x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right)S{u}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.\hfill \end{array}$
(1.11)

They proved that if the sequences $\left\{{\alpha }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ of parameters satisfy appropriate conditions, then the sequence $\left\{{x}_{n}\right\}$ generated by (1.11) converges strongly to the unique solution of the variational inequality

$〈\left(A-\gamma f\right)z,x-z〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in F\left(S\right)\cap EP\left(F\right),$
(1.12)

which is the optimality condition for the minimization problem

$\underset{x\in F\left(S\right)\cap EP\left(F\right)}{min}\frac{1}{2}〈Ax,x〉-h\left(x\right),$

where h is a potential function for γf (i.e., ${h}^{\prime }\left(x\right)=\gamma f\left(x\right)$ for $x\in H$).

Let ${T}_{1},{T}_{2},\dots$ be an infinite sequence of mappings of C into itself, and let ${\lambda }_{1},{\lambda }_{2},\dots$ be real numbers such that $0\le {\lambda }_{i}\le 1$ for every $i\in \mathbb{N}$. Then for any $n\in \mathbb{N}$, Takahashi  (see ) defined a mapping ${W}_{n}$ of C into itself as follows: (1.13)

Such a mapping ${W}_{n}$ is called the W-mapping generated by ${T}_{n},{T}_{n-1},\dots ,{T}_{1}$ and ${\lambda }_{n},{\lambda }_{n-1},\dots ,{\lambda }_{1}$.

Recently, using process (1.13), Yao et al.  proved the following result.

Theorem 1.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $F:C×C\to \mathbb{R}$ be an equilibrium bifunction satisfying the conditions:

1. (1)

F is monotone, that is, $F\left(x,y\right)+F\left(y,x\right)\le 0$ for all $x,y\in C$;

2. (2)

for each $x,y,z\in C$, ${lim}_{t\to 0}F\left(tz+\left(1-t\right)x,y\right)\le F\left(x,y\right)$;

3. (3)

for each $x\in C$, $y↦F\left(x,y\right)$ is convex and lower semicontinuous.

Let ${\left\{{T}_{i}\right\}}_{i=1}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into C such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)\ne \mathrm{\varnothing }$. Suppose $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ are three sequences in $\left(0,1\right)$ such that ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$ and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$. Suppose the following conditions are satisfied:

1. (1)
${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$

and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (2)
$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$

;

3. (3)
${lim inf}_{n\to \mathrm{\infty }}{r}_{n}>0$

and ${lim}_{n\to \mathrm{\infty }}\left({r}_{n+1}-{r}_{n}\right)=0$.

Let f be a contraction of H into itself, and let ${x}_{0}\in H$ be given arbitrarily. Then the sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ generated iteratively by

$\left\{\begin{array}{c}F\left({y}_{n},x\right)+\frac{1}{{r}_{n}}〈x-{y}_{n},{y}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C,\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}{W}_{n}{y}_{n},\hfill \end{array}$

converge strongly to ${x}^{\ast }\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)$, the unique solution of the minimization problem

$\underset{x\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)}{min}\frac{1}{2}{\parallel x\parallel }^{2}-h\left(x\right),$

where h is a potential function for f.

Very recently, using process (1.13), Chen  proved the following result.

Theorem 1.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\left\{{T}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from C to C such that the common fixed point set $\mathrm{\Omega }={\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)\ne \mathrm{\varnothing }$. Let $f:C\to H$ be an α-contraction, and let $A:H\to H$ be a self-adjoint, strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$. Let σ be a constant such that $0<\gamma \alpha <\overline{\gamma }$. For an arbitrary initial point ${x}_{0}$ belonging to C, one defines a sequence ${\left\{{x}_{n}\right\}}_{n\ge 0}$ iteratively

${x}_{n+1}={P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{x}_{n}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(1.14)

where $\left\{{\alpha }_{n}\right\}$ is a real sequence in $\left[0,1\right]$. Assume the sequence $\left\{{\alpha }_{n}\right\}$ satisfies the following conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$.

Then the sequence $\left\{{x}_{n}\right\}$ generated by (1.14) converges in norm to the unique solution ${x}^{\ast }$, which solves the following variational inequality:

${x}^{\ast }\in \mathrm{\Omega }\phantom{\rule{1em}{0ex}}\mathit{\text{such that}}\phantom{\rule{0.1em}{0ex}}〈\left(A-\gamma f\right){x}^{\ast },{x}^{\ast }-\stackrel{ˆ}{x}〉\ge 0,\mathrm{\forall }\stackrel{ˆ}{x}\in \mathrm{\Omega }.$
(1.15)

Motivated by this result, we introduce the following explicit general iterative scheme:

$\left\{\begin{array}{c}{x}_{1}\in H,\hfill \\ F\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in H,\hfill \\ {x}_{n+1}={P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$
(1.16)

where ${\left\{{T}_{n}\right\}}_{n\in \mathbb{N}}$ is a family of nonexpansive mappings from H into itself such that ${\bigcap }_{n\in \mathbb{N}}F\left({T}_{n}\right)$ is nonempty, $F:C×C\to \mathbb{R}$ is an equilibrium bifunction, A is a strongly positive operator on H, f is a contraction of H into itself with $\alpha \in \left(0,1\right)$, $\left\{{\alpha }_{n}\right\}$, $\left\{{r}_{n}\right\}$, $\left\{{\lambda }_{n}\right\}$ suitable sequences in and $\left\{{W}_{n}\right\}$ is the sequence of a W-mapping generated by ${\left\{{T}_{n}\right\}}_{n\in \mathbb{N}}$ and $\left\{{\lambda }_{n}\right\}$. Let U be defined by $Ux={lim}_{n\to \mathrm{\infty }}{W}_{n}x={lim}_{n\to \mathrm{\infty }}{U}_{n,1}x$ for every $x\in C$ using process (1.13). We shall prove under mild conditions that $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ strongly converge to a point ${x}^{\ast }\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)$, which is the unique solution of the variational inequality

$〈\left(A-\gamma f\right){x}^{\ast },{x}^{\ast }-\stackrel{ˆ}{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }\stackrel{ˆ}{x}\in \bigcap _{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right),$
(1.17)

or, equivalently, the unique solution of the minimization problem

$\underset{x\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)}{min}\frac{1}{2}〈A\stackrel{ˆ}{x},\stackrel{ˆ}{x}〉-h\left(\stackrel{ˆ}{x}\right),$

where h is a potential function for γf.

## 2 Preliminaries

Let H be a real Hilbert space with the norm $\parallel \cdot \parallel$ and the inner product $〈\cdot ,\cdot 〉$, and let C be a closed convex subset of H. We call $f:C\to H$ an α-contraction if there exists a constant $\alpha \in \left[0,1\right)$ such that

$\parallel f\left(x\right)-f\left(y\right)\parallel \le \alpha \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Let A be a strongly positive linear bounded operator on a Hilbert space H with a constant $\overline{\gamma }$; that is, there exists $\overline{\gamma }>0$ such that

$〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H.$

Next, we denote weak convergence and strong convergence by notations and →, respectively. A space X is said to satisfy Opial’s condition  if for each sequence $\left\{{x}_{n}\right\}$ in X which converges weakly to a point $x\in X$, we have

$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-x\parallel <\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in X,y\ne x.$

For every point $x\in H$, there exists a unique nearest point in C, denoted by ${P}_{C}x$, such that

$\parallel x-{P}_{C}x\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
${P}_{C}$

is called the (nearest point or metric) projection of H onto C. In addition, ${P}_{C}x$ is characterized by the following properties: ${P}_{C}x\in C$ and (2.1) (2.2)

Recall that a mapping $T:H\to H$ is said to be firmly nonexpansive mapping if

${\parallel Tx-Ty\parallel }^{2}\le 〈Tx-Ty,x-y〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in H.$

It is well known that ${P}_{C}$ is a firmly nonexpansive mapping of H onto C and satisfies

${\parallel {P}_{C}x-{P}_{C}y\parallel }^{2}\le 〈x-y,{P}_{C}x-{P}_{C}y〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in H.$
(2.3)

If A is an α-inverse-strongly monotone mapping of C into H, then it is obvious that A is $\frac{1}{\alpha }$-Lipschitz continuous. We also have that for all $x,y\in C$ and $\lambda >0$,

$\begin{array}{rcl}{\parallel \left(I-\lambda A\right)x-\left(I-\lambda A\right)y\parallel }^{2}& =& {\parallel x-y-\lambda \left(Ax-Ay\right)\parallel }^{2}\\ =& {\parallel x-y\parallel }^{2}-2\lambda 〈Ax-Ay,x-y〉+{\lambda }^{2}{\parallel Ax-Ay\parallel }^{2}\\ \le & {\parallel x-y\parallel }^{2}+\lambda \left(\lambda -2\alpha \right){\parallel Ax-Ay\parallel }^{2}.\end{array}$
(2.4)

So, if $\lambda \le 2\alpha$, then $I-\lambda A$ is a nonexpansive mapping of C into H.

The following lemmas will be useful for proving the convergence result of this paper.

Lemma 2.1 Let H be a real Hilbert space. Then for all $x,y\in H$,

1. (1)
${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,x+y〉$

;

2. (2)
${\parallel x+y\parallel }^{2}\ge {\parallel x\parallel }^{2}+2〈y,x〉$

.

Lemma 2.2 ()

Let $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ be bounded sequences in a Banach space X, and let $\left\{{\beta }_{n}\right\}$ be a sequence in $\left[0,1\right]$ with $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$. Suppose that ${x}_{n+1}=\left(1-{\beta }_{n}\right){y}_{n}+{\beta }_{n}{x}_{n}$ for all integers $n\ge 0$ and ${lim sup}_{n\to \mathrm{\infty }}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel {y}_{n}-{x}_{n}\parallel =0$.

Lemma 2.3 ()

Assume that $F:C×C\to \mathbb{R}$, let us assume that F satisfies the following conditions:

(A1) $F\left(x,x\right)=0$ for all $x\in C$;

(A2) F is monotone, i.e., $F\left(x,y\right)+F\left(y,x\right)\le 0$ for all $x,y\in C$;

(A3) for each $x,y,z\in C$, ${lim}_{t\to 0}F\left(tz+\left(1-t\right)x,y\right)\le F\left(x,y\right)$;

(A4) for each $x\in C$, $y↦F\left(x,y\right)$ is convex and lower semicontinuous.

Lemma 2.4 ()

Assume that $F:C×C\to \mathbb{R}$ satisfies (A1)-(A4). For $r>0$ and $x\in H$, define a mapping ${T}_{r}:H\to C$ as follows:

${T}_{r}\left(x\right)=\left\{z\in C:F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\}$

for all $z\in H$. Then the following hold:

1. 1.
${T}_{r}$

is single-valued;

2. 2.
${T}_{r}$

is firmly nonexpansive, i.e., for any $x,y\in H$,

${\parallel {T}_{r}x-{T}_{r}y\parallel }^{2}\le 〈{T}_{r}x-{T}_{r}y,x-y〉;$
3. 3.
$F\left({T}_{r}\right)=EP\left(F\right)$

;

4. 4.

EP(F) is closed and convex.

Lemma 2.5 ()

Let H be a Hilbert space, C be a closed convex subset of H, and $S:C\to C$ be a nonexpansive mapping with $F\left(S\right)\ne \mathrm{\varnothing }$. If $\left\{{x}_{n}\right\}$ is a sequence in C weakly converging to $x\in C$ and if $\left\{\left(I-S\right){x}_{n}\right\}$ converges strongly to y, then $\left(I-S\right)x=y$.

Lemma 2.6 ()

Assume that $\left\{{a}_{n}\right\}$ is a sequence of nonnegative real numbers such that

${a}_{n+1}\le \left(1-{\gamma }_{n}\right){a}_{n}+{\delta }_{n},\phantom{\rule{1em}{0ex}}n\ge 0,$

where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence in such that

1. (1)
${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$

;

2. (2)
${lim sup}_{n\to \mathrm{\infty }}\frac{{\delta }_{n}}{{\gamma }_{n}}\le 0$

or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n}|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{a}_{n}=0$.

Lemma 2.7 ()

Let H be a Hilbert space, C be a nonempty closed convex subset of H, and $f:H\to H$ be a contraction with a coefficient $0<\alpha <1$, and let A be a strongly positive linear bounded operator with a coefficient $\overline{\gamma }>0$. Then, for $0<\gamma <\frac{\overline{\gamma }}{\alpha }$,

$〈x-y,\left(A-\gamma f\right)x-\left(A-\gamma f\right)y〉\ge \left(\overline{\gamma }-\gamma \alpha \right){\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}x,y\in H.$

That is, $A-\gamma f$ is strongly monotone with a coefficient $\overline{\gamma }-\gamma \alpha$.

Lemma 2.8 ()

Assume A is a strongly positive linear bounded operator on a Hilbert space H with a coefficient $\overline{\gamma }>0$ and $0<\rho \le {\parallel A\parallel }^{-1}$. Then $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$.

Lemma 2.9 ( and )

Let C be a nonempty closed convex subset of a Banach space E. Let ${\left\{{T}_{i}\right\}}_{i=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings of C into itself with ${\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$, and let ${\left\{{\lambda }_{i}\right\}}_{i=1}^{\mathrm{\infty }}$ be a real sequence such that $0<{\lambda }_{i}\le b<1$, $\mathrm{\forall }i\ge 1$. Then:

1. (1)
${W}_{n}$

is nonexpansive and $F\left({W}_{n}\right)={\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)$ for each $n\ge 1$;

2. (2)

for each $x\in C$ and for each positive integer k, the ${lim}_{n\to \mathrm{\infty }}{U}_{n,k}x$ exists;

3. (3)

the mapping $U:C\to C$ defined by

$Ux=\underset{n\to \mathrm{\infty }}{lim}{W}_{n}x=\underset{n\to \mathrm{\infty }}{lim}{U}_{n,1}x,\phantom{\rule{1em}{0ex}}x\in C$

is a nonexpansive mapping satisfying $F\left(U\right)={\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{i}\right)$ and it is called the W-mapping generated by ${T}_{1},{T}_{2},\dots$ and ${\lambda }_{1},{\lambda }_{2},\dots$ ;

1. (4)
${lim}_{m,n\to \mathrm{\infty }}{sup}_{x\in K}\parallel {W}_{m}x-{W}_{n}x\parallel =0$

for any bounded subset K of E.

## 3 Main results

In this section, we introduce our algorithm and prove its strong convergence.

Theorem 3.1 Let C be a closed convex subset of a real Hilbert space H. Let F be a bifunction from $H×H$ into satisfying (A1)-(A4). Let f be a contraction of H into itself with $\alpha \in \left(0,1\right)$, and let ${T}_{n}$ be a sequence of nonexpansive mappings of C into itself such that $\mathrm{\Omega }={\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)\cap EP\left(F\right)\ne \mathrm{\varnothing }$. Let $A:H\to H$ be a strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$ with $0<\gamma <\frac{\overline{\gamma }}{\alpha }$. Let ${\lambda }_{1},{\lambda }_{2},\dots$ be a sequence of real numbers such that $0<{\lambda }_{n}\le b<1$ for every $n=1,2,\dots$ . Let ${W}_{n}$ be a W-mapping of C into itself generated by ${T}_{n},{T}_{n-1},\dots ,{T}_{1}$ and ${\lambda }_{n},{\lambda }_{n-1},\dots ,{\lambda }_{1}$. Let U be defined by $Ux={lim}_{n\to \mathrm{\infty }}{W}_{n}x={lim}_{n\to \mathrm{\infty }}{U}_{n,1}x$ for every $x\in C$. Let $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by ${x}_{1}\in H$ and

$\left\{\begin{array}{c}F\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in H,\hfill \\ {x}_{n+1}={P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$
(3.1)

where $\left\{{\alpha }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{r}_{n}\right\}$ is a sequence in $\left[0,\mathrm{\infty }\right)$. Suppose that $\left\{{\alpha }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

(C3) ${lim}_{n\to \mathrm{\infty }}{r}_{n}=r>0$.

Then both $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ converge strongly to ${x}^{\ast }\in \mathrm{\Omega }$, which is the unique solution of the variational inequality

$〈\left(A-\gamma f\right){x}^{\ast },{x}^{\ast }-\stackrel{ˆ}{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }\stackrel{ˆ}{x}\in \mathrm{\Omega }.$
(3.2)

Equivalently, one has ${x}^{\ast }={P}_{\mathrm{\Omega }}\left(I-A+\gamma f\right)\left({x}^{\ast }\right)$.

Proof We observe that ${P}_{\mathrm{\Omega }}\left(\gamma f+\left(I-A\right)\right)$ is a contraction. Indeed, for all $x,y\in H$, we have

$\begin{array}{rcl}\parallel {P}_{\mathrm{\Omega }}\left(\gamma f+\left(I-A\right)\right)\left(x\right)-{P}_{\mathrm{\Omega }}\left(\gamma f+\left(I-A\right)\right)\left(y\right)\parallel & \le & \parallel \left(\gamma f+\left(I-A\right)\right)\left(x\right)-\left(\gamma f+\left(I-A\right)\right)\left(y\right)\parallel \\ \le & \gamma \parallel f\left(x\right)-f\left(y\right)\parallel +\parallel I-A\parallel \parallel x-y\parallel \\ \le & \gamma \alpha \parallel x-y\parallel +\left(1-\overline{\gamma }\right)\parallel x-y\parallel \\ <& \left(1-\left(\overline{\gamma }-\gamma \alpha \right)\right)\parallel x-y\parallel .\end{array}$

Banach’s contraction mapping principle guarantees that ${P}_{\mathrm{\Omega }}\left(\gamma f+\left(I-A\right)\right)$ has a unique fixed point, say ${x}^{\ast }\in H$. That is, ${x}^{\ast }={P}_{\mathrm{\Omega }}\left(\gamma f+\left(I-A\right)\right)\left({x}^{\ast }\right)$. Note that by Lemma 2.4, we can write

${x}_{n+1}={P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{T}_{{r}_{n}}{x}_{n}\right],$

where

${T}_{{r}_{n}}\left(x\right)=\left\{z\in H:F\left(z,y\right)+\frac{1}{{r}_{n}}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in H\right\}.$

Moreover, since ${\alpha }_{n}\to 0$ as $n\to \mathrm{\infty }$ by condition (C1), we assume that ${\alpha }_{n}\le {\parallel A\parallel }^{-1}$ for all $n\in \mathbb{N}$. From Lemma 2.8, we know that if $0<\rho <{\parallel A\parallel }^{-1}$, then $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$. We divide the proof into seven steps as follows.

Step 1. Show that the sequences $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ are bounded.

Let $\stackrel{ˆ}{x}\in \mathrm{\Omega }$. Then $\stackrel{ˆ}{x}\in EP\left(F\right)$. From Lemma 2.4, we have

$\parallel {u}_{n}-\stackrel{ˆ}{x}\parallel =\parallel {T}_{{r}_{n}}{x}_{n}-{T}_{{r}_{n}}\stackrel{ˆ}{x}\parallel \le \parallel {x}_{n}-\stackrel{ˆ}{x}\parallel .$

Thus, we have

$\begin{array}{rcl}\parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel & =& \parallel {P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]-\stackrel{ˆ}{x}\parallel \\ \le & \parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}-\stackrel{ˆ}{x}\parallel \\ \le & {\alpha }_{n}\gamma \parallel f\left({x}_{n}\right)-f\left(\stackrel{ˆ}{x}\right)\parallel +\parallel I-{\alpha }_{n}A\parallel \parallel {W}_{n}{u}_{n}-\stackrel{ˆ}{x}\parallel +{\alpha }_{n}\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel \\ \le & {\alpha }_{n}\gamma \alpha \parallel {x}_{n}-\stackrel{ˆ}{x}\parallel +\left(1-{\alpha }_{n}\overline{\gamma }\right)\parallel {u}_{n}-\stackrel{ˆ}{x}\parallel +{\alpha }_{n}\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel \\ =& \left(1-{\alpha }_{n}\left(\overline{\gamma }-\gamma \alpha \right)\right)\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel +{\alpha }_{n}\left(\overline{\gamma }-\gamma \alpha \right)\frac{\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel }{\left(\overline{\gamma }-\gamma \alpha \right)}.\end{array}$

By induction, we have

$\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel \le max\left\{\parallel {x}_{1}-\stackrel{ˆ}{x}\parallel ,\frac{\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel }{\overline{\gamma }-\gamma \alpha }\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.$

This shows that the sequence $\left\{{x}_{n}\right\}$ is bounded, so are $\left\{{u}_{n}\right\}$, $\left\{f\left({x}_{n}\right)\right\}$ and $\left\{{W}_{n}{u}_{n}\right\}$.

Step 2. Show that $\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$.

Let $\stackrel{ˆ}{x}\in \mathrm{\Omega }$. Since ${T}_{i}$ and ${U}_{n,i}$ are nonexpansive and ${T}_{i}\stackrel{ˆ}{x}=\stackrel{ˆ}{x}={U}_{n,i}\stackrel{ˆ}{x}$ for every $n\in \mathbb{N}$ and $i\le n+1$, it follows that

$\begin{array}{rcl}\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel & =& \parallel {\lambda }_{1}{T}_{1}{U}_{n+1,2}{u}_{n}-{\lambda }_{1}{T}_{1}{U}_{n,2}{u}_{n}\parallel \\ \le & {\lambda }_{1}\parallel {U}_{n+1,2}{u}_{n}-{U}_{n,2}{u}_{n}\parallel \\ =& {\lambda }_{1}\parallel {\lambda }_{2}{T}_{2}{U}_{n+1,3}{u}_{n}-{\lambda }_{2}{T}_{2}{U}_{n,3}{u}_{n}\parallel \\ \le & {\lambda }_{1}{\lambda }_{2}\parallel {U}_{n+1,3}{u}_{n}-{U}_{n,3}{u}_{n}\parallel \\ ⋮\\ \le & \left(\prod _{i=1}^{n}{\lambda }_{i}\right)\parallel {U}_{n+1,n+1}{u}_{n}-\stackrel{ˆ}{x}+\stackrel{ˆ}{x}-{U}_{n,n+1}{u}_{n}\parallel \\ \le & \left(\prod _{i=1}^{n}{\lambda }_{i}\right)\left(\parallel {U}_{n+1,n+1}{u}_{n}-\stackrel{ˆ}{x}+\stackrel{ˆ}{x}-{U}_{n,n+1}{u}_{n}\parallel \right)\\ \le & 2\left(\prod _{i=1}^{n}{\lambda }_{i}\right)\parallel {u}_{n}-\stackrel{ˆ}{x}\parallel .\end{array}$

Since $\left\{{u}_{n}\right\}$ is bounded and $0<{\lambda }_{n}\le b<1$ for any $n\in \mathbb{N}$, the following holds:

$\underset{n\to \mathrm{\infty }}{lim}\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel =0.$

Step 3. Show that $\parallel {x}_{n+1}-{x}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$.

Setting $S=2{P}_{C}-I$, we have S is nonexpansive. Note that ${W}_{n}=\left(1-{\lambda }_{1}\right)I+{\lambda }_{1}{T}_{1}{U}_{n,2}$. Then we can write

$\begin{array}{rcl}{x}_{n+1}& =& \frac{I+S}{2}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]\\ =& \frac{1-{\alpha }_{n}}{2}{W}_{n}{u}_{n}+\frac{{\alpha }_{n}}{2}\left(\gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}+{W}_{n}{u}_{n}\right)\\ +\frac{1}{2}S\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]\\ =& \frac{1-{\alpha }_{n}}{2}\left[\left(1-{\lambda }_{1}\right)I+{\lambda }_{1}{T}_{1}{U}_{n,2}\right]{u}_{n}+\frac{{\alpha }_{n}}{2}\left(\gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}+{W}_{n}{u}_{n}\right)\\ +\frac{1}{2}S\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]\\ =& \frac{\left(1-{\lambda }_{1}\right)\left(1-{\alpha }_{n}\right)}{2}{u}_{n}+\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{2}{T}_{1}{U}_{n,2}{u}_{n}+\frac{{\alpha }_{n}}{2}\left(\gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}+{W}_{n}{u}_{n}\right)\\ +\frac{1}{2}S\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right].\end{array}$
(3.3)

Note that

$0<\underset{n\to \mathrm{\infty }}{lim}\frac{\left(1-{\lambda }_{1}\right)\left(1-{\alpha }_{n}\right)}{2}=\frac{1-{\lambda }_{1}}{2}<1,$

and

$\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{2}+\frac{1}{2}=\frac{1+{\lambda }_{1}}{2}-\frac{{\lambda }_{1}}{2}{\alpha }_{n}.$

From (3.3), we have

$\begin{array}{rcl}{x}_{n+1}& =& \left[1-\left(\frac{1+{\lambda }_{1}}{2}+\frac{1-{\lambda }_{1}}{2}{\alpha }_{n}\right)\right]{u}_{n}+\left(\frac{1+{\lambda }_{1}}{2}+\frac{1-{\lambda }_{1}}{2}{\alpha }_{n}\right)\\ ×\left(\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{2}{T}_{1}{U}_{n,2}{u}_{n}+\frac{{\alpha }_{n}}{2}\left(\gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}+{W}_{n}{u}_{n}\right)\\ +\frac{1}{2}S\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]\right)/\left(\frac{1+{\lambda }_{1}}{2}+\frac{1-{\lambda }_{1}}{2}{\alpha }_{n}\right)\\ =& \left[1-\left(\frac{1+{\lambda }_{1}}{2}+\frac{1-{\lambda }_{1}}{2}{\alpha }_{n}\right)\right]{u}_{n}+\left(\frac{1+{\lambda }_{1}}{2}+\frac{1-{\lambda }_{1}}{2}{\alpha }_{n}\right){y}_{n},\end{array}$
(3.4)

where

$\begin{array}{rcl}{y}_{n}& =& \left(\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{2}{T}_{1}{U}_{n,2}{u}_{n}+\frac{{\alpha }_{n}}{2}\left(\gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}+{W}_{n}{u}_{n}\right)\\ +\frac{1}{2}S\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]\right)/\left(\frac{1+{\lambda }_{1}}{2}+\frac{1-{\lambda }_{1}}{2}{\alpha }_{n}\right)\\ =& \left({\lambda }_{1}\left(1-{\alpha }_{n}\right){T}_{1}{U}_{n,2}{u}_{n}+{\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}+{W}_{n}{u}_{n}\right)\\ +S\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]\right)/\left(1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}\right).\end{array}$

Set ${e}_{n}=\gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}+{W}_{n}{u}_{n}$ and ${d}_{n}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}$ for all n. Then

${y}_{n}=\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right){T}_{1}{U}_{n,2}{u}_{n}+{\alpha }_{n}{e}_{n}+S{d}_{n}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.$

It follows that

$\begin{array}{rcl}{y}_{n+1}-{y}_{n}& =& \frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right){T}_{1}{U}_{n+1,2}{u}_{n+1}+{\alpha }_{n+1}{e}_{n+1}+S{d}_{n+1}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\\ -\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right){T}_{1}{U}_{n,2}{u}_{n}+{\alpha }_{n}{e}_{n}+S{d}_{n}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\\ =& \frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\left({T}_{1}{U}_{n+1,2}{u}_{n+1}-{T}_{1}{U}_{n,2}{u}_{n}\right)\\ +\left(\frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\right){T}_{1}{U}_{n,2}{u}_{n}\\ +\frac{{\alpha }_{n+1}{e}_{n+1}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{{\alpha }_{n}{e}_{n}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\\ +\frac{S{d}_{n+1}-S{d}_{n}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}+\left(\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\right)S{d}_{n}.\end{array}$

Thus,

$\begin{array}{rcl}\parallel {y}_{n+1}-{y}_{n}\parallel & \le & \frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\parallel {T}_{1}{U}_{n+1,2}{u}_{n+1}-{T}_{1}{U}_{n,2}{u}_{n}\parallel \\ +|\frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}|\parallel {T}_{1}{U}_{n,2}{u}_{n}\parallel \\ +\frac{{\alpha }_{n+1}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\parallel {e}_{n+1}\parallel +\frac{{\alpha }_{n}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\parallel {e}_{n}\parallel \\ +\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\parallel S{d}_{n+1}-S{d}_{n}\parallel \\ +|\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}|\parallel S{d}_{n}\parallel .\end{array}$

Since S is nonexpansive, we obtain that

$\begin{array}{rcl}\parallel S{d}_{n+1}-S{d}_{n}\parallel & \le & \parallel {d}_{n+1}-{d}_{n}\parallel \\ =& \parallel {\alpha }_{n+1}\gamma f\left({x}_{n+1}\right)+\left(I-{\alpha }_{n+1}A\right){W}_{n+1}{u}_{n+1}-\left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right)\parallel \\ \le & {\alpha }_{n+1}\parallel \gamma f\left({x}_{n+1}\right)-A{W}_{n+1}{u}_{n+1}\parallel +{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}\right)\parallel \\ +\parallel {W}_{n+1}{u}_{n+1}-{W}_{n}{u}_{n}\parallel \\ \le & {\alpha }_{n+1}\parallel \gamma f\left({x}_{n+1}\right)-A{W}_{n+1}{u}_{n+1}\parallel +{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}\right)\parallel \\ +\parallel {W}_{n+1}{u}_{n+1}-{W}_{n+1}{u}_{n}\parallel +\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel \\ \le & {\alpha }_{n+1}\parallel \gamma f\left({x}_{n+1}\right)-A{W}_{n+1}{u}_{n+1}\parallel +{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}\right)\parallel +\parallel {u}_{n+1}-{u}_{n}\parallel .\\ +\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel .\end{array}$

Since ${T}_{i}$ and ${U}_{n,i}$ are nonexpansive, we have

$\begin{array}{rcl}\parallel {T}_{1}{U}_{n+1,2}{u}_{n}-{T}_{1}{U}_{n,2}{u}_{n}\parallel & \le & \parallel {U}_{n+1,2}{u}_{n}-{U}_{n,2}{u}_{n}\parallel \\ =& \parallel {\lambda }_{2}{T}_{2}{U}_{n+1,3}{u}_{n}-{\lambda }_{2}{T}_{2}{U}_{n,3}{u}_{n}\parallel \\ \le & {\lambda }_{2}\parallel {U}_{n+1,3}{u}_{n}-{U}_{n,3}{u}_{n}\parallel \\ \le & \cdots \\ \le & {\lambda }_{2}\cdots {\lambda }_{n}\parallel {U}_{n+1,n+1}{u}_{n}-{U}_{n,n+1}{u}_{n}\parallel \\ \le & M\prod _{i=2}^{n}{\lambda }_{i},\end{array}$

where $M>0$ is a constant such that $\parallel {U}_{n+1,n+1}{u}_{n}-{U}_{n,n+1}{u}_{n}\parallel \le M$ for all $n\ge 0$. So,

$\begin{array}{rcl}\parallel {T}_{1}{U}_{n+1,2}{u}_{n+1}-{T}_{1}{U}_{n,2}{u}_{n}\parallel & \le & \parallel {T}_{1}{U}_{n+1,2}{u}_{n+1}-{T}_{1}{U}_{n+1,2}{u}_{n}\parallel +\parallel {T}_{1}{U}_{n+1,2}{u}_{n}-{T}_{1}{U}_{n,2}{u}_{n}\parallel \\ \le & \parallel {u}_{n+1}-{u}_{n}\parallel +M\prod _{i=2}^{n}{\lambda }_{i}.\end{array}$

Hence,

$\begin{array}{rcl}\parallel {y}_{n+1}-{y}_{n}\parallel & \le & \frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\parallel {u}_{n+1}-{u}_{n}\parallel +M\prod _{i=2}^{n}{\lambda }_{i}\\ +|\frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}|\parallel {T}_{1}{U}_{n,2}{u}_{n}\parallel \\ +\frac{{\alpha }_{n+1}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\parallel {e}_{n+1}\parallel +\frac{{\alpha }_{n}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\parallel {e}_{n}\parallel \\ +\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\left({\alpha }_{n+1}\parallel \gamma f\left({x}_{n+1}\right)-A{W}_{n+1}{u}_{n+1}\parallel \\ +{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}\parallel +\parallel {u}_{n+1}-{u}_{n}\parallel \right)+\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel \\ +|\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}|\parallel S{d}_{n}\parallel \\ =& \frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\parallel {u}_{n+1}-{u}_{n}\parallel \\ +|\frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}|\parallel {T}_{1}{U}_{n,2}{u}_{n}\parallel \\ +\frac{{\alpha }_{n+1}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\parallel {e}_{n+1}\parallel +\frac{{\alpha }_{n}}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\parallel {e}_{n}\parallel \\ +\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}\left({\alpha }_{n+1}\parallel \gamma f\left({x}_{n+1}\right)-A{W}_{n+1}{u}_{n+1}\parallel \\ +{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-A{W}_{n}{u}_{n}\parallel +\parallel {u}_{n+1}-{u}_{n}\parallel \right)+\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel \\ +|\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}|\parallel S{d}_{n}\parallel .\end{array}$

Note that: (1)

By condition (C1), we have

$\frac{{\lambda }_{1}\left(1-{\alpha }_{n+1}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{{\lambda }_{1}\left(1-{\alpha }_{n}\right)}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\to 0$

and

$\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_{1}+\left(1-{\lambda }_{1}\right){\alpha }_{n}}\to 0.$
1. (2)
$\parallel {W}_{n+1}{u}_{n}-{W}_{n}{u}_{n}\parallel \to 0$

as $n\to \mathrm{\infty }$ because of Step 2.

Therefore,

$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {u}_{n+1}-{u}_{n}\parallel \right)\le 0.$

By Lemma 2.2, we get

$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{u}_{n}\parallel =0.$

Hence, from (3.4), we deduce

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1+{\lambda }_{1}}{2}+\frac{1-{\lambda }_{1}}{2}{\alpha }_{n}\right)\parallel {y}_{n}-{u}_{n}\parallel =0.$
(3.5)

Step 4. Show that $\parallel {x}_{n}-{W}_{n}{u}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$. Indeed, we have

$\begin{array}{rcl}\parallel {x}_{n}-{W}_{n}{u}_{n}\parallel & \le & \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{W}_{n}{u}_{n}\parallel \\ =& \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]-{W}_{n}{u}_{n}\parallel \\ \le & \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}-{W}_{n}{u}_{n}\parallel \\ =& \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel -{\alpha }_{n}A{W}_{n}{u}_{n}+{\alpha }_{n}\gamma f\left({x}_{n}\right)\parallel \\ \le & \parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\left(\parallel A\parallel \parallel {W}_{n}{u}_{n}\parallel +\gamma \parallel f\left({x}_{n}\right)\parallel \right).\end{array}$

Then

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{W}_{n}{u}_{n}\parallel \le \underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\left(\parallel A\parallel \parallel {W}_{n}{u}_{n}\parallel +\gamma \parallel f\left({x}_{n}\right)\parallel \right)=0.$
(3.6)

Thus, from (3.6), we obtain

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{W}_{n}{u}_{n}\parallel =0.$

Step 5. Show that $\parallel {x}_{n}-{u}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$.

Let $\stackrel{ˆ}{x}\in \mathrm{\Omega }$. Since ${T}_{{r}_{n}}$ is firmly nonexpansive, it follows

$\begin{array}{rcl}{\parallel \stackrel{ˆ}{x}-{u}_{n}\parallel }^{2}& =& {\parallel {T}_{{r}_{n}}\stackrel{ˆ}{x}-{T}_{{r}_{n}}{x}_{n}\parallel }^{2}\\ \le & 〈{T}_{{r}_{n}}{x}_{n}-{T}_{{r}_{n}}\stackrel{ˆ}{x},{x}_{n}-\stackrel{ˆ}{x}〉\\ \le & 〈{T}_{{r}_{n}}{x}_{n}-\stackrel{ˆ}{x},{x}_{n}-\stackrel{ˆ}{x}〉\\ =& 〈{u}_{n}-\stackrel{ˆ}{x},{x}_{n}-\stackrel{ˆ}{x}〉\\ =& \frac{1}{2}\left({\parallel {u}_{n}-\stackrel{ˆ}{x}\parallel }^{2}+{\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}-{\parallel {x}_{n}-{u}_{n}\parallel }^{2}\right).\end{array}$

Then

${\parallel {u}_{n}-\stackrel{ˆ}{x}\parallel }^{2}\le {\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}-{\parallel {x}_{n}-{u}_{n}\parallel }^{2}.$

Since we have

$\begin{array}{rcl}{\parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel }^{2}& =& {\parallel {P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}\right]-{P}_{C}\stackrel{ˆ}{x}\parallel }^{2}\\ =& {\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}-\stackrel{ˆ}{x}\parallel }^{2}\\ =& {\parallel \left(I-{\alpha }_{n}A\right)\left({W}_{n}{u}_{n}-\stackrel{ˆ}{x}\right)+{\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-A\stackrel{ˆ}{x}\right)\parallel }^{2}\\ \le & {\left(1-{\alpha }_{n}\overline{\gamma }\right)}^{2}{\parallel {W}_{n}{u}_{n}-\stackrel{ˆ}{x}\parallel }^{2}+2{\alpha }_{n}〈\gamma f\left({x}_{n}\right)-A\stackrel{ˆ}{x},{x}_{n+1}-\stackrel{ˆ}{x}〉\\ \le & {\left(1-{\alpha }_{n}\overline{\gamma }\right)}^{2}{\parallel {u}_{n}-\stackrel{ˆ}{x}\parallel }^{2}+2{\alpha }_{n}\gamma 〈f\left({x}_{n}\right)-f\left(\stackrel{ˆ}{x}\right),{x}_{n+1}-\stackrel{ˆ}{x}〉\\ +2{\alpha }_{n}〈\gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x},{x}_{n+1}-\stackrel{ˆ}{x}〉\\ \le & {\left(1-{\alpha }_{n}\overline{\gamma }\right)}^{2}\left({\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}-{\parallel {x}_{n}-{u}_{n}\parallel }^{2}\right)+2{\alpha }_{n}\gamma \alpha \parallel {x}_{n}-\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel \\ +2{\alpha }_{n}\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel \\ =& \left(1-2{\alpha }_{n}\overline{\gamma }+{\left({\alpha }_{n}\overline{\gamma }\right)}^{2}\right){\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}-{\left(1-{\alpha }_{n}\overline{\gamma }\right)}^{2}{\parallel {x}_{n}-{u}_{n}\parallel }^{2}\\ +2{\alpha }_{n}\gamma \alpha \parallel {x}_{n}-\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel +2{\alpha }_{n}\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel \\ \le & {\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}+{\alpha }_{n}{\overline{\gamma }}^{2}{\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}-{\left(1-{\alpha }_{n}\overline{\gamma }\right)}^{2}{\parallel {x}_{n}-{u}_{n}\parallel }^{2}\\ +2{\alpha }_{n}\gamma \alpha \parallel {x}_{n}-\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel +2{\alpha }_{n}\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel ,\end{array}$

and hence

$\begin{array}{rcl}{\left(1-{\alpha }_{n}\overline{\gamma }\right)}^{2}{\parallel {x}_{n}-{u}_{n}\parallel }^{2}& \le & {\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}-{\parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel }^{2}+{\alpha }_{n}{\overline{\gamma }}^{2}{\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}\\ +2{\alpha }_{n}\gamma \alpha \parallel {x}_{n}-\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel +2{\alpha }_{n}\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel \\ \le & \parallel {x}_{n}-{x}_{n+1}\parallel \left(\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel \right)+{\alpha }_{n}{\overline{\gamma }}^{2}{\parallel {x}_{n}-\stackrel{ˆ}{x}\parallel }^{2}\\ +2{\alpha }_{n}\gamma \alpha \parallel {x}_{n}-\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel +2{\alpha }_{n}\parallel \gamma f\left(\stackrel{ˆ}{x}\right)-A\stackrel{ˆ}{x}\parallel \parallel {x}_{n+1}-\stackrel{ˆ}{x}\parallel .\end{array}$

Therefore, we have $\parallel {x}_{n}-{u}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$.

Step 6. Show that ${lim sup}_{n\to \mathrm{\infty }}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n}-{x}^{\ast }〉\le 0$.

We can choose a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that

$\underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{{n}_{i}}-{x}^{\ast }〉=\underset{n\to \mathrm{\infty }}{lim sup}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n}-{x}^{\ast }〉.$

Let

$A\left({x}_{{n}_{i}}\right)=\left\{x\in H:\underset{i\to \mathrm{\infty }}{lim sup}\parallel {x}_{{n}_{i}}-x\parallel =\underset{y\in H}{inf}\underset{i\to \mathrm{\infty }}{lim sup}\parallel {x}_{{n}_{i}}-y\parallel \right\}$

be the asymptotic center of $\left\{{x}_{{n}_{i}}\right\}$. Since $\left\{{x}_{{n}_{i}}\right\}$ is bounded and H is a Hilbert space, it is well known that $A\left({x}_{{n}_{i}}\right)$ is a singleton; say $A\left({x}_{{n}_{i}}\right)=\left\{\stackrel{˜}{x}\right\}$. Set

$L=\underset{i\in \mathbb{N}}{sup}\parallel \gamma f\left({x}_{{n}_{i}}\right)-A{W}_{{n}_{i}}{u}_{{n}_{i}}\parallel$

and for every $x\in H$ define

$Wx=\underset{i\to \mathrm{\infty }}{lim}{W}_{{n}_{i}}x$
(3.7)

and

${T}_{r}\left(x\right)=\left\{z\in H:F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in H\right\}.$

Note that

$\begin{array}{rcl}\parallel {x}_{{n}_{i}}-W\stackrel{˜}{x}\parallel & \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {x}_{{n}_{i}+1}-W\stackrel{˜}{x}\parallel \\ =& \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {P}_{C}\left[{\alpha }_{{n}_{i}}\gamma f\left({x}_{{n}_{i}}\right)+\left(I-{\alpha }_{{n}_{i}}A\right){W}_{{n}_{i}}{u}_{{n}_{i}}\right]-W\stackrel{˜}{x}\parallel \\ \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {\alpha }_{{n}_{i}}\gamma f\left({x}_{{n}_{i}}\right)+\left(I-{\alpha }_{{n}_{i}}A\right){W}_{{n}_{i}}{u}_{{n}_{i}}-W\stackrel{˜}{x}\parallel \\ =& \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {W}_{{n}_{i}}{u}_{{n}_{i}}-W\stackrel{˜}{x}+{\alpha }_{{n}_{i}}\left(\gamma f\left({x}_{{n}_{i}}\right)-A{W}_{{n}_{i}}{u}_{{n}_{i}}\right)\parallel \\ \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {W}_{{n}_{i}}{u}_{{n}_{i}}-{W}_{{n}_{i}}{x}_{{n}_{i}}\parallel +\parallel {W}_{{n}_{i}}{x}_{{n}_{i}}-{W}_{{n}_{i}}\stackrel{˜}{x}\parallel \\ +\parallel {W}_{{n}_{i}}\stackrel{˜}{x}-W\stackrel{˜}{x}\parallel +{\alpha }_{{n}_{i}}L\\ \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {u}_{{n}_{i}}-{x}_{{n}_{i}}\parallel +\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel +\parallel {W}_{{n}_{i}}\stackrel{˜}{x}-W\stackrel{˜}{x}\parallel +{\alpha }_{{n}_{i}}L.\end{array}$

By Steps 1-5, condition (C1) and (3.7), we derive

$\begin{array}{rcl}\underset{i\to \mathrm{\infty }}{lim sup}\parallel {x}_{{n}_{i}}-W\stackrel{˜}{x}\parallel & \le & \underset{i\to \mathrm{\infty }}{lim sup}\parallel {u}_{{n}_{i}}-{x}_{{n}_{i}}\parallel +\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel +\parallel {W}_{{n}_{i}}\stackrel{˜}{x}-W\stackrel{˜}{x}\parallel \\ \le & \underset{i\to \mathrm{\infty }}{lim sup}\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel .\end{array}$

That is, $W\stackrel{˜}{x}\in A\left({x}_{{n}_{i}}\right)$. Therefore $W\stackrel{˜}{x}=\stackrel{˜}{x}$. Next, we show that $\stackrel{˜}{x}={T}_{r}\stackrel{˜}{x}$.

Note that for any $x\in H$ and $a,b>0$, we have

$F\left({T}_{a}x,y\right)+\frac{1}{a}〈y-{T}_{a}x,{T}_{a}x-x〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in H$

and

$F\left({T}_{b}x,y\right)+\frac{1}{b}〈y-{T}_{b}x,{T}_{b}x-x〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in H,$

then

$F\left({T}_{a}x,{T}_{b}x\right)+\frac{1}{a}〈{T}_{b}x-{T}_{a}x,{T}_{a}x-x〉\ge 0$

and

$F\left({T}_{b}x,{T}_{a}x\right)+\frac{1}{b}〈{T}_{a}x-{T}_{b}x,{T}_{a}x-x〉\ge 0.$

Summing up the last inequalities and using (A2), we obtain

$〈{T}_{a}x-{T}_{b}x,\frac{{T}_{b}x-x}{b}-\frac{{T}_{a}x-x}{a}〉\ge 0.$

Hence we have

$\begin{array}{rcl}0& \le & 〈{T}_{a}x-{T}_{b}x,{T}_{b}x-x-\frac{b}{a}\left({T}_{a}x-x\right)〉\\ =& 〈{T}_{a}x-{T}_{b}x,{T}_{b}x-{T}_{a}x+{T}_{a}x-x-\frac{b}{a}\left({T}_{a}x-x\right)〉\\ =& 〈{T}_{a}x-{T}_{b}x,\left({T}_{b}x-{T}_{a}x\right)+\left(1-\frac{b}{a}\right)\left({T}_{a}x-x\right)〉\\ \le & -{\parallel {T}_{a}x-{T}_{b}x\parallel }^{2}+|1-\frac{b}{a}|\parallel {T}_{a}x-{T}_{b}x\parallel \left(\parallel {T}_{a}x\parallel +\parallel x\parallel \right).\end{array}$

We derive then

$\parallel {T}_{a}x-{T}_{b}x\parallel \le \frac{|b-a|}{a}\left(\parallel {T}_{a}x\parallel +\parallel x\parallel \right).$

It follows that

$\begin{array}{rcl}\parallel {x}_{{n}_{i}}-{T}_{r}\stackrel{˜}{x}\parallel & \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {x}_{{n}_{i}+1}-{T}_{r}\stackrel{˜}{x}\parallel \\ =& \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {P}_{C}\left[{\alpha }_{{n}_{i}}\gamma f\left({x}_{{n}_{i}}\right)+\left(I-{\alpha }_{{n}_{i}}A\right){W}_{{n}_{i}}{u}_{{n}_{i}}\right]-{T}_{r}\stackrel{˜}{x}\parallel \\ \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {\alpha }_{{n}_{i}}\gamma f\left({x}_{{n}_{i}}\right)+\left(I-{\alpha }_{{n}_{i}}A\right){W}_{{n}_{i}}{u}_{{n}_{i}}-{T}_{r}\stackrel{˜}{x}\parallel \\ =& \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {W}_{{n}_{i}}{u}_{{n}_{i}}-{T}_{r}\stackrel{˜}{x}+{\alpha }_{{n}_{i}}\left(\gamma f\left({x}_{{n}_{i}}\right)-A{W}_{{n}_{i}}{u}_{{n}_{i}}\right)\parallel \\ \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {W}_{{n}_{i}}{u}_{{n}_{i}}-{x}_{{n}_{i}}\parallel +\parallel {T}_{{r}_{{n}_{i}}}{x}_{{n}_{i}}-{T}_{{r}_{{n}_{i}}}\stackrel{˜}{x}\parallel +\parallel {x}_{{n}_{i}}-{T}_{{r}_{{n}_{i}}}{x}_{{n}_{i}}\parallel \\ +\parallel {T}_{{r}_{{n}_{i}}}\stackrel{˜}{x}-{T}_{r}\stackrel{˜}{x}\parallel +{\alpha }_{{n}_{i}}L\\ \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {u}_{{n}_{i}}-{x}_{{n}_{i}}\parallel +\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel +\parallel {x}_{{n}_{i}}-{u}_{{n}_{i}}\parallel \\ +\parallel {T}_{{r}_{{n}_{i}}}\stackrel{˜}{x}-{T}_{r}\stackrel{˜}{x}\parallel +{\alpha }_{{n}_{i}}L\\ \le & \parallel {x}_{{n}_{i}+1}-{x}_{{n}_{i}}\parallel +\parallel {u}_{{n}_{i}}-{x}_{{n}_{i}}\parallel +\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel +\parallel {x}_{{n}_{i}}-{u}_{{n}_{i}}\parallel \\ +\frac{|{r}_{{n}_{i}}-r|}{r}\left(\parallel {T}_{r}\stackrel{˜}{x}\parallel +\parallel \stackrel{˜}{x}\parallel \right)+{\alpha }_{{n}_{i}}L.\end{array}$

By Steps 2-5, conditions (C1) and (C3), we obtain

$\underset{i\to \mathrm{\infty }}{lim sup}\parallel {x}_{{n}_{i}}-{T}_{r}\stackrel{˜}{x}\parallel \le \underset{i\to \mathrm{\infty }}{lim sup}\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel$

and $\stackrel{˜}{x}={T}_{r}\stackrel{˜}{x}$. Thus $\stackrel{˜}{x}\in F\left(W\right)\cap F\left({T}_{r}\right)=\mathrm{\Omega }$ by Lemma 2.3 and 2.9. Fix $t\in \left(0,1\right)$, $x\in H$ and set $y=\stackrel{˜}{x}+tx$. Then

${\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}-tx\parallel }^{2}\le {\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel }^{2}+2t〈x,\stackrel{˜}{x}+tx-{x}_{{n}_{i}}〉.$

By the minimizing property of $\stackrel{˜}{x}$ and since ${\parallel \cdot \parallel }^{2}$ is continuous and increasing in $\left[0,\mathrm{\infty }\right)$, we have

$\begin{array}{rcl}\underset{i\to \mathrm{\infty }}{lim sup}{\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel }^{2}& \le & \underset{i\to \mathrm{\infty }}{lim sup}{\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}-tx\parallel }^{2}\\ \le & \underset{i\to \mathrm{\infty }}{lim sup}{\parallel {x}_{{n}_{i}}-\stackrel{˜}{x}\parallel }^{2}+2t\underset{i\to \mathrm{\infty }}{lim sup}〈x,\stackrel{˜}{x}+tx-{x}_{{n}_{i}}〉.\end{array}$

Thus,

$\underset{i\to \mathrm{\infty }}{lim sup}〈x,\stackrel{˜}{x}+tx-{x}_{{n}_{i}}〉\ge 0.$

On the other hand,

$〈x,\stackrel{˜}{x}-{x}_{{n}_{i}}〉=〈x,\stackrel{˜}{x}+tx-{x}_{{n}_{i}}〉-t{\parallel x\parallel }^{2}.$

Hence we obtain

$\underset{i\to \mathrm{\infty }}{lim sup}〈x,\stackrel{˜}{x}-{x}_{{n}_{i}}〉=\underset{t\to 0}{lim}\left(\underset{i\to \mathrm{\infty }}{lim sup}〈x,\stackrel{˜}{x}+tx-{x}_{{n}_{i}}〉-t{\parallel x\parallel }^{2}\right)\ge 0.$

Set $x=\gamma f\left({x}^{\ast }\right)-A{x}^{\ast }$. Since $\stackrel{˜}{x}\in \mathrm{\Omega }$, we obtain

$\begin{array}{rcl}0& \le & \underset{i\to \mathrm{\infty }}{lim sup}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },\stackrel{˜}{x}-{x}_{{n}_{i}}〉\\ \le & 〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },\stackrel{˜}{x}-{x}^{\ast }〉+\underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}^{\ast }-{x}_{{n}_{i}}〉\\ \le & \underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}^{\ast }-{x}_{{n}_{i}}〉.\end{array}$

So that

$\underset{n\to \mathrm{\infty }}{lim sup}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n}-{x}^{\ast }〉=-\underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{{n}_{i}}-{x}^{\ast }〉\le 0.$

Step 7. Show that both $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ strongly converge to ${x}^{\ast }\in \mathrm{\Omega }$, which is the unique solution of the variational inequality (3.2). Indeed, we note that

${\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}=〈{x}_{n+1}-{d}_{n},{x}_{n+1}-{x}^{\ast }〉+〈{d}_{n}-{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉.$

Since $〈{x}_{n+1}-{d}_{n},{x}_{n+1}-{x}^{\ast }〉\le 0$, we get

$\begin{array}{rcl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}& \le & 〈{d}_{n}-{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉\\ =& 〈{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{u}_{n}-{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉\\ =& 〈{\alpha }_{n}\gamma f\left({x}_{n}\right)-{\alpha }_{n}\gamma f\left({x}^{\ast }\right)+{W}_{n}{u}_{n}-{\alpha }_{n}A{W}_{n}{u}_{n}-{x}^{\ast }\\ +{\alpha }_{n}A{x}^{\ast }+{\alpha }_{n}\gamma f\left({x}^{\ast }\right)-{\alpha }_{n}A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉\\ =& 〈{\alpha }_{n}\gamma \left(f\left({x}_{n}\right)-f\left({x}^{\ast }\right)\right)+\left(I-{\alpha }_{n}A\right)\left({W}_{n}{u}_{n}-{x}^{\ast }\right),{x}_{n+1}-{x}^{\ast }〉\\ +{\alpha }_{n}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉\\ \le & \left({\alpha }_{n}\gamma \parallel f\left({x}_{n}\right)-f\left({x}^{\ast }\right)\parallel +\parallel I-{\alpha }_{n}A\parallel \parallel {W}_{n}{u}_{n}-{x}^{\ast }\parallel \right)\parallel {x}_{n+1}-{x}^{\ast }\parallel \\ +{\alpha }_{n}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉\\ \le & \left(1-{\alpha }_{n}\left(\overline{\gamma }-\gamma \alpha \right)\right)\parallel {x}_{n}-{x}^{\ast }\parallel \parallel {x}_{n+1}-{x}^{\ast }\parallel +{\alpha }_{n}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉\\ \le & \left(\frac{{\left[1-{\alpha }_{n}\left(\overline{\gamma }-\gamma \alpha \right)\right]}^{2}}{2}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\frac{1}{2}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\\ +{\alpha }_{n}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉.\end{array}$

It then follows that

${\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le \left[1-{\alpha }_{n}\left(\overline{\gamma }-\gamma \alpha \right)\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+2{\alpha }_{n}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉.$
(3.8)

Let ${a}_{n}={\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}$, ${\gamma }_{n}={\alpha }_{n}\left(\overline{\gamma }-\gamma \alpha \right)$ and ${\delta }_{n}=2{\alpha }_{n}〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉$.

Then, we can write the last inequality as

${a}_{n+1}\le \left(1-{\gamma }_{n}\right){a}_{n}+{\delta }_{n}.$

Note that in virtue of condition (C2), ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$. Moreover,

$\underset{n\to \mathrm{\infty }}{lim sup}\frac{{\delta }_{n}}{{\gamma }_{n}}=\frac{1}{\overline{\gamma }-\gamma \alpha }\underset{n\to \mathrm{\infty }}{lim sup}2〈\gamma f\left({x}^{\ast }\right)-A{x}^{\ast },{x}_{n+1}-{x}^{\ast }〉.$

By Step 5, we obtain

$\underset{n\to \mathrm{\infty }}{lim sup}\frac{{\delta }_{n}}{{\gamma }_{n}}\le 0.$
(3.9)

Now, applying Lemma 2.6 to (3.8), we conclude that ${x}_{n}\to {x}^{\ast }$ as $n\to \mathrm{\infty }$. Furthermore, since $\parallel {u}_{n}-{x}^{\ast }\parallel =\parallel {T}_{{r}_{n}}{x}_{n}-{T}_{{r}_{n}}{x}^{\ast }\parallel \le \parallel {x}_{n}-{x}^{\ast }\parallel$, we then have that ${u}_{n}\to {x}^{\ast }$ as $n\to \mathrm{\infty }$. The proof is now complete. □

Setting $A\equiv I$ and $\gamma =1$ in Theorem 3.1, we have the following result.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $F:C×C\to \mathbb{R}$ be an equilibrium bifunction satisfying the conditions:

1. (1)

F is monotone, that is, $F\left(x,y\right)+F\left(y,x\right)\le 0$ for all $x,y\in C$;

2. (2)

for each $x,y,z\in C$, ${lim}_{t\to 0}F\left(tz+\left(1-t\right)x,y\right)\le F\left(x,y\right)$;

3. (3)

for each $x\in C$, $y↦F\left(x,y\right)$ is convex and lower semicontinuous.

Let ${\left\{{T}_{i}\right\}}_{i=1}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into C such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)\ne \mathrm{\varnothing }$. Suppose $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ satisfy the following conditions:

1. (1)
${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$

and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (2)
${lim inf}_{n\to \mathrm{\infty }}{r}_{n}>0$

and ${lim}_{n\to \mathrm{\infty }}\left({r}_{n+1}-{r}_{n}\right)=0$.

Let f be a contraction of C into itself, and let ${x}_{0}\in H$ be given arbitrarily. Then the sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ generated iteratively by

$\left\{\begin{array}{c}F\left({y}_{n},x\right)+\frac{1}{{r}_{n}}〈x-{y}_{n},{y}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C,\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){W}_{n}{y}_{n},\hfill \end{array}$

converge strongly to ${x}^{\ast }\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)$, the unique solution of the minimization problem

$\underset{x\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({T}_{i}\right)\cap EP\left(F\right)}{min}\frac{1}{2}{\parallel x\parallel }^{2}-h\left(x\right),$

where h is a potential function for f.

Setting $F=0$ in Theorem 3.1, we have the following result.

Corollary 3.3 ()

Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\left\{{T}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from C to C such that the common fixed point set $\mathrm{\Omega }={\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)\ne \mathrm{\varnothing }$. Let $f:C\to H$ be an α-contraction and $A:H\to H$ be a strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$. Let γ be a constant such that $0<\gamma \alpha <\overline{\gamma }$. For an arbitrary initial point ${x}_{0}$ belonging to C, one defines a sequence ${\left\{{x}_{n}\right\}}_{n\ge 0}$ iteratively

${x}_{n+1}={P}_{C}\left[{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right){W}_{n}{x}_{n}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(3.10)

where $\left\{{\alpha }_{n}\right\}$ is a real sequence in $\left[0,1\right]$. Assume that the sequence $\left\{{\alpha }_{n}\right\}$ satisfies the following conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$.

Then the sequence $\left\{{x}_{n}\right\}$ generated by (3.10) converges in norm to the unique solution ${x}^{\ast }$, which solves the following variational inequality:

${x}^{\ast }\in \mathrm{\Omega }\phantom{\rule{1em}{0ex}}\mathit{\text{such that}}\phantom{\rule{0.1em}{0ex}}〈\left(A-\gamma f\right){x}^{\ast },{x}^{\ast }-\stackrel{ˆ}{x}〉\le 0,\mathrm{\forall }\stackrel{ˆ}{x}\in \mathrm{\Omega }.$
(3.11)

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## Acknowledgements

The author would like to thank Asst. Prof. Dr. Rabian Wangkeeree for his useful suggestions and the referees for their valuable comments and suggestions.

## Author information

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Correspondence to Kiattisak Rattanaseeha.

### Competing interests

The author declares that they have no competing interests.

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Rattanaseeha, K. The general iterative methods for equilibrium problems and fixed point problems of a countable family of nonexpansive mappings in Hilbert spaces. J Inequal Appl 2013, 153 (2013). https://doi.org/10.1186/1029-242X-2013-153

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• DOI: https://doi.org/10.1186/1029-242X-2013-153

### Keywords

• equilibrium problem
• fixed point
• nonexpansive mapping
• variational inequality
• strongly positive operator
• Hilbert spaces 